Question

If the solubility product of lead iodide (PbI₂) is 3.2 x 10^-8 . Then its solubility in moles / litre will be ……

(a) 2 x 10^-3

(b) 4 x 10^-4 

(c) 1.6 x 10^-5 

(d) 1.8 x 10^-5

Answer 1

(a) 2 x 10^-3

K_sp = 4s³ 

4s³ = 3.2 x 10^-8

s = 2 x 10^-3M